\documentclass[a4paper]{article}
\usepackage[affil-it]{authblk}
\usepackage[backend=bibtex,style=numeric]{biblatex}
\usepackage[UTF8]{ctex}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\geometry{margin=1.5cm, vmargin={0pt,1cm}}
\setlength{\topmargin}{-1cm}
\setlength{\paperheight}{29.7cm}
\setlength{\textheight}{25.3cm}

\addbibresource{citation.bib}

\begin{document}
% =================================================
\title{Numerical Analysis homework \# 2}

\author{王劼 Wang Jie 3220100105
  \thanks{Electronic address: \texttt{2645443470@qq.com}}}
\affil{(math), Zhejiang University }


\date{Due time: \today}

\maketitle

\begin{abstract}
    theoretical homework 2  
\end{abstract}





% ============================================
\section*{theoretical homework}

Complete all the theoretical homework for Section 2.11.1. Please submit your homework 
in PDF format. You are encouraged to use LaTeX to complete your homework. 
If the format of your homework or your text is not neat enough, it may affect your score.
\cite{wangheyu2024}

\subsection*{Question 2.11.1 I}  


\textbf{Answer:} \\

\textbf{1. Determine $\xi(x)$ explicitly:}

Given \( f(x) = \frac{1}{x} \), we first compute \( f''(x) \):
\[
f'(x) = -\frac{1}{x^2},
\]
\[
f''(x) = \frac{2}{x^3}.
\]
Then we know \( p_{1}(f; x) = \frac{3}{2} - \frac{x}{2} \) by \( x_0=1, x_1=2 \) and \( p_{1}(f; x_0)=f(x_0), p_{1}(f; x_1)=f(x_1) \).

Substituting \( f''(x) \) into the given formula:
\[
f(x) - p_{1}(f; x) = \frac{f''(\xi(x))}{2}(x - x_{0})(x - x_{1}),
\]
we have:
\[
\frac{1}{x} - \left(\frac{3 - x}{2}\right) = \frac{1}{\xi(x)^3}(x - 1)(x - 2).
\]
Simplifying the left-hand side:
\[
\frac{2 - x(3 - x)}{2x} = \frac{1}{\xi(x)^3}(x - 1)(x - 2).
\]
This reduces to:
\[
\frac{(x - 1)(x - 2)}{2x} = \frac{1}{\xi(x)^3}(x - 1)(x - 2).
\]

Since \( x - 1 \neq 0 \) and \( x - 2 \neq 0 \) within \( (1, 2) \), we can divide both sides by \( (x - 1)(x - 2) \):
\[
\frac{1}{2x} = \frac{1}{\xi(x)^3}.
\]
Thus, we find:
\[
\xi(x)^3 = 2x,
\]
and therefore:
\[
\xi(x) = \sqrt[3]{2x}.
\]

\textbf{2. Extend the domain of $\xi$ continuously from $(x_0, x_1)$ to $[x_0, x_1]$:}

We need to find a continuous extension of \( \xi(x) \) on \( [1, 2] \). Since \( \xi(x) \) is already defined on \( (1, 2) \), we only need to consider the endpoints \( x = 1 \) and \( x = 2 \).

As \( x \to 1^+ \), \( \xi(x) \to \sqrt[3]{2} \).
As \( x \to 2^- \), \( \xi(x) \to \sqrt[3]{4} \).

Thus, we can define \( \xi(x) \) as:
\[
\xi(x) = 
\begin{cases} 
\sqrt[3]{2} & \text{if } x = 1, \\
\sqrt[3]{2x} & \text{if } 1 < x < 2, \\
\sqrt[3]{4} & \text{if } x = 2.
\end{cases}
\]
So, we have:
\[
\xi(x) = \sqrt[3]{2x} \quad \text{for } x \in [1, 2].
\]

\textbf{3. Find $\max \xi(x)$, $\min \xi(x)$, and $\max f''(\xi(x))$:}

The minimum of \( \xi(x) \) is \( \sqrt[3]{2} \), achieved at \( x = 1 \).
The maximum of \( \xi(x) \) is \( \sqrt[3]{4} \), achieved at \( x = 2 \).
We compute \( f''(\xi(x)) \):
\[
f''(\xi(x)) = \frac{2}{\xi(x)^3},
\]
and since the range of \( \xi(x) \) is \( [\sqrt[3]{2}, \sqrt[3]{4}] \), the maximum of \( f''(\xi(x)) \) is achieved at \( \xi(x) = \sqrt[3]{2} \), which gives us:
\[
f''(\sqrt[3]{2}) = 1.
\]

\subsection*{Question 2.11.1 II} 

\textbf{Answer:} \\

\textbf{1. Constructing the Lagrange Interpolation Polynomial:}

   Using Lagrange interpolation, for \( n + 1 \) points \( (x_0, f_0), (x_1, f_1), \ldots, (x_n, f_n) \), we can define the polynomial \( p_1(x) \) as follows:
   \[
   p_1(x) = \sum_{i=0}^{n} f_i \cdot L_i(x),
   \]
   where \( L_i(x) \) is the Lagrange basis polynomial given by:
   \[
   L_i(x) = \prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j}.
   \]
   The polynomial \( p_1(x) \) constructed this way is of degree \( n \) and satisfies \( p_1(x_i) = f_i \).

\textbf{2. Ensuring Non-negativity:}

   To ensure that \( p(x) \) is non-negative, we can make the following adjustments. Since \( f_i \geq 0 \), we can construct a polynomial in a non-negative form. Consider a new polynomial:
   \[
   p(x) = \left( \sum_{i=0}^{n} \sqrt{f_i} \cdot L_i(x) \right)^2.
   \]

This polynomial satisfies \( p(x_i) = f_i \) for each \( i \), is non-negative for all real numbers \( x \) and it's degree is 2n, equaling to $p \in P_{2n}^+$.

By combining Lagrange interpolation with the square form, we successfully constructed a polynomial that meets the conditions of the problem while ensuring non-negativity. 

\subsection*{Question 2.11.1 III} 

\textbf{Answer:}\\ 
To prove the statement 

\[
\forall t \in \mathbb{R}, \quad f[t, t+1, \ldots, t+n] = \frac{(e-1)^n}{n!} e^t,
\]

where \( f(x) = e^x \), we will use mathematical induction.

\textbf{1. Base Case (\( n = 0 \))}

When \( n = 0 \), the interpolating polynomial \( f[t] \) has only one point \( t \). Thus we have:

\[
f[t] = f(t) = e^t.
\]

The right-hand side evaluates to:

\[
\frac{(e-1)^0}{0!} e^t = 1 \cdot e^t = e^t.
\]

Thus, the base case holds.

\textbf{2. Induction Hypothesis}

Assume that the statement holds for some \( n = k \):

\[
f[t, t+1, \ldots, t+k] = \frac{(e-1)^k}{k!} e^t.
\]

\textbf{3. Induction Step}

Now we need to prove that it holds for \( n = k + 1 \):

\[
f[t, t+1, \ldots, t+k+1] = \frac{f[t+1, t+2, \ldots, t+k+1] - f[t, t+1, \ldots, t+k]}{t+k+1 - t} = \frac{\frac{(e-1)^k}{k!} e^{t+1} - \frac{(e-1)^k}{k!} e^t}{k+1} = \frac{(e-1)^{k+1}}{(k+1)!} e^t.
\]

Therefore, it can be concluded by mathematical induction that 
\[
\forall t \in \mathbb{R}, \quad f[t, t+1, \ldots, t+n] = \frac{(e-1)^n}{n!} e^t,
\].

\textbf{4. Determining the Position of \(\xi\)}

From the interpolation result, we have

\[
f[0, 1, \ldots, n] = \frac{1}{n!} f^{(n)}(\xi) = \frac{1}{n!} e^{\xi},
\]

which we can relate to the previous expression for \( f[t, t+1, \ldots, t+n] \) when $t = 0$:

\[
\frac{(e-1)^n}{n!} = \frac{1}{n!} e^\xi.
\]

From this, we obtain:

\[
e^{\xi} = (e-1)^n.
\]

then, we have:

\[
\xi = n\log(e-1).
\]

To compare \( \log(e - 1) \) and 0.5, we can evaluate \( \log(e - 1) \).

\[
\log(e - 1) \approx 0.5413
\]

Since \( 0.5413 > 0.5 \), we conclude that:

\[
\xi = n\log(e-1) > n/2
\]

\(\xi\) is located to the right of the midpoint \( n/2 \).

\subsection*{Question 2.11.1 IV}

\textbf{Answer:}\\

\textbf{1. construct \( p_3(f; x) \)}

We will use Newton's interpolation method to construct the interpolation polynomial \( p_3(f; x) \) through the known points \( f(0)=5 \), \( f(1)=3 \), \( f(3)=5 \), and \( f(4)=12 \).

Let \( x_0 = 0 \), \( x_1 = 1 \), \( x_2 = 3 \), \( x_3 = 4 \), and the corresponding function values be \( f(x_0) = 5 \), \( f(x_1) = 3 \), \( f(x_2) = 5 \), \( f(x_3) = 12 \).

\begin{enumerate}
    \item \textbf{Zeroth-order divided differences:}
    \[
    f[x_0] = f(0) = 5
    \]
    \[
    f[x_1] = f(1) = 3
    \]
    \[
    f[x_2] = f(3) = 5
    \]
    \[
    f[x_3] = f(4) = 12
    \]

    \item \textbf{First-order divided differences:}
    \[
    f[x_0, x_1] = \frac{f(1) - f(0)}{1 - 0} = \frac{3 - 5}{1} = -2
    \]
    \[
    f[x_1, x_2] = \frac{f(3) - f(1)}{3 - 1} = \frac{5 - 3}{2} = 1
    \]
    \[
    f[x_2, x_3] = \frac{f(4) - f(3)}{4 - 3} = \frac{12 - 5}{1} = 7
    \]

    \item \textbf{Second-order divided differences:}
    \[
    f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{3 - 0} = \frac{1 - (-2)}{3} = \frac{3}{3} = 1
    \]
    \[
    f[x_1, x_2, x_3] = \frac{f[x_2, x_3] - f[x_1, x_2]}{4 - 1} = \frac{7 - 1}{3} = \frac{6}{3} = 2
    \]

    \item \textbf{Third-order divided differences:}
    \[
    f[x_0, x_1, x_2, x_3] = \frac{f[x_1, x_2, x_3] - f[x_0, x_1, x_2]}{4 - 0} = \frac{2 - 1}{4} = \frac{1}{4}
    \]
\end{enumerate}

Using the Newton interpolation formula:
\[
p_3(x) = f[x_0] + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) + f[x_0, x_1, x_2, x_3](x - x_0)(x - x_1)(x - x_2)
\]

Substituting in the known values:
\[
p_3(x) = 5 - 2(x - 0) + 1(x - 0)(x - 1) + \frac{1}{4}(x - 0)(x - 1)(x - 3)
\]

Simplifying each term:
\[
p_3(x) = 5 - 2x + (x^2 - x) + \frac{1}{4}(x^3 - 4x^2 + 3x)
\]
\[
= 5 - 2x + x^2 - x + \frac{1}{4}x^3 - x^2 + \frac{3}{4}x
\]
\[
= 5 - 3x + \frac{3}{4}x + \frac{1}{4}x^3
\]
\[
= 5 - \frac{9}{4}x + \frac{1}{4}x^3
\]

\textbf{2. Find the Minimum Value}

Next, we need to find the minimum value of this polynomial. First, we differentiate:
\[
p_3'(x) = -\frac{9}{4} + \frac{3}{4}x^2
\]
Setting the derivative to zero:
\[
-\frac{9}{4} + \frac{3}{4}x^2 = 0
\]
\[
\frac{3}{4}x^2 = \frac{9}{4}
\]
\[
x^2 = 3
\]
\[
x = \sqrt{3} \approx 1.7321
\]

Thus, the approximate value of \( x_{\min} \) is 
\[
\boxed{1.7321}
\]

\subsection*{Question 2.11.1 V}

\textbf{Answer:}\\

\textbf(1. calculate the divided difference $ f[0, 1, 1, 1, 2, 2] $)

To calculate the divided difference $ f[0, 1, 1, 1, 2, 2] $ for the function $ f(x) = x^7 $:

\begin{enumerate}
   \item \textbf{Zeroth-order divided differences:}
   \[
   f[0] = f(0) = 0
   \]
   \[
   f[1] = f(1) = 1
   \]
   \[
   f[2] = f(2) = 128
   \]

   \item \textbf{Calculate \( f[1, 1, 1] \):}
   \[
   f[1, 1, 1] = \frac{f^(2)(1)}{2!} = 21
   \]

   \item \textbf{Calculate \( f[1, 1, 2] \):}
   \[
   f[1, 1, 2] = \frac{f[1, 2] - f[1, 1]}{2 - 1}
   \]
   Calculating \( f[1, 2] \):
   \[
   f[1, 2] = \frac{f(2) - f(1)}{2 - 1} = \frac{2^7 - 1^7}{1} = 128 - 1 = 127
   \]
   Calculating \( f[1, 1] \):
   \[
   f[1, 1] = \frac{f^(1)(1)}{1!} = 7
   \]

   Thus, we have:
   \[
   f[1, 1, 2] = 120
   \]

   \item \textbf{Calculate \( f[0, 1, 1] \):}
   \[
   f[0, 1, 1] = \frac{f[1, 1] - f[0, 1]}{1 - 0}
   \]
   We already found \( f[1, 1] = 7 \), now calculate \( f[0, 1] \):
   \[
   f[0, 1] = \frac{f(1) - f(0)}{1 - 0} = \frac{1^7 - 0^7}{1} = 1
   \]
   Therefore:
   \[
   f[0, 1, 1] = \frac{7 - 1}{1} = 6
   \]

   \item \textbf{Calculate \( f[0, 1, 1, 1, 2] \):}
   \[
   f[0, 1, 1, 1, 2] = \frac{f[1, 1, 1, 2] - f[0, 1, 1, 1]}{2 - 0}
   \]
   We calculate \( f[1, 1, 1, 2] \):
   \[
   f[1, 1, 1, 2] = \frac{f[1, 1, 2] - f[1, 1, 1]}{2 - 1} = 99
   \]
   We calculate \( f[0, 1, 1, 1] \):
   \[
   f[0, 1, 1, 1] = \frac{f[1, 1, 1] - f[0, 1, 1]}{1 - 0} = 15
   \]
   Now substituting these values:
   \[
   f[0, 1, 1, 1, 2] = \frac{99 - 15}{2} = 42
   \]

   \item \textbf{Calculate \( f[0, 1, 1, 1, 2, 2] \):}
   \[
   f[0, 1, 1, 1, 2, 2] = \frac{f[1, 1, 1, 2, 2] - f[0, 1, 1, 1, 2]}{2 - 0}
   \]
   Calculate \( f[1, 1, 2, 2] \):
   \[
   f[1, 1, 2, 2] = \frac{f[1, 2, 2] - f[1, 1, 2]}{2 - 1} = 201
   \]
   Calculate \( f[1, 1, 1, 2, 2] \):
   \[
   f[1, 1, 1, 2, 2] = \frac{f[1, 1, 2, 2] - f[1, 1, 1, 2]}{2 - 1} = \frac{201 - 99}{2 - 1} = 102
   \]
   Substituting the values:
   \[
   f[0, 1, 1, 1, 2, 2] = \frac{102 - 42}{2} = 30
   \]
\end{enumerate}

\textbf{2. Finding the Point \( \xi \)}

According to the property of divided differences, the repeated divided difference can be expressed in terms of the function's derivatives at some point \( \xi \). For the repeated fifth divided difference, we have:
\[
f[0, 1, 1, 1, 2, 2] = \frac{f^{(5)}(\xi)}{5!}
\]

Here, \( f^{(5)}(x) = 7 \times 6 \times 5 \times 4 \times 3 x^{2} = 2520 x^{2} \). Thus, we can write:
\[
30 = \frac{2520 \xi^2}{120}
\]
Simplifying gives:
\[
30 = 21 \xi^2 \quad \Rightarrow \quad \xi^2 = \frac{10}{7} = \sqrt{\frac{10}{7}} \approx 1.195
\]

We conclude that:
\[
f[0, 1, 1, 1, 2, 2] = 30 \quad \text{and} \quad \xi \approx 1.195 \in (0, 2).
\]

\subsection*{Question 2.11.1 VI}

\textbf{Answer:}\\

\textbf{1. construct \( p_4(f; x) \) to estimate f(2) by Hermite interpolation}

\begin{enumerate}
   \item \textbf{Zeroth-order divided differences:}
   \[
   f[0] = f(0) = 1
   \]
   \[
   f[1] = f(1) = 2
   \]
   \[
   f[3] = f(3) = 0
   \]

   \item \textbf{First-order divided differences:}
   \[
   f[0, 1] = \frac{f(1) - f(0)}{1 - 0} = 1
   \]
   \[
   f[1, 1] = \frac{f(3) - f(1)}{3 - 1} = -1
   \]
   \[
   f[1, 3] = \frac{f(4) - f(3)}{4 - 3} = -1
   \]
   \[
   f[3, 3] = \frac{f(4) - f(3)}{4 - 3} = 0
   \]


   \item \textbf{Second-order divided differences:}
   \[
   f[0, 1, 1] = \frac{f[1, 1] - f[0, 1]}{1 - 0} = -2
   \]
   \[
   f[1, 1, 3] = \frac{f[1, 3] - f[1, 1]}{3 - 1} = 0
   \]
   \[
   f[1, 3, 3] = \frac{f[3, 3] - f[1, 3]}{3 - 1} = 1/2
   \]

   \item \textbf{Third-order divided differences:}
   \[
   f[0, 1, 1, 3] = \frac{f[1, 1, 3] - f[0, 1, 1]}{3 - 0} = 2/3
   \]
   \[
   f[1, 1, 3, 3] = \frac{f[1, 3, 3] - f[1, 1, 3]}{3 - 1} = 1/4
   \]

   \item \textbf{Fourth-order divided differences:}
   \[
   f[0, 1, 1, 3, 3] = \frac{f[1, 1, 3, 3] - f[0, 1, 1, 3]}{3 - 0} = -5/36
   \]
\end{enumerate}

Thus, we obtain:
\[
p_4(f; x) = 1 + 1*(x-0) - 2*(x-0)(x-1) + \frac{2}{3}*(x-0)(x-1)^2 - \frac{5}{36}(x-0)(x-1)^2(x-3).
\]

The approximate value of \( f(2) \) is $p_4(f; 2) = \frac{11}{18} \approx 0.611$

\textbf{2. Error Estimation}

Given that \( f \in \mathcal{C}^5[0,3] \) and \(\left| f^{(5)}(x) \right| \leq M\) in \([0,3]\), the error in Hermite interpolation is given by:

\[
E(x) = \frac{f^{(5)}(\xi)}{5!} \prod_{i=0}^{n} (x - x_i)^2,
\]

for some \( \xi \in [0,3] \). Since \( x = 2 \) is in the interval and \( x_0 = 0, x_1 = 1, x_2 = 3 \):

\[
E(2) \leq \frac{M}{120} \cdot (2 - 0) \cdot (2 - 1)^2 \cdot (2 - 3)^2 = \frac{M}{120} \cdot 2 \cdot 1 \cdot 1 = \frac{M}{60}.
\]

Thus, the maximum possible error in estimating \( f(2) \) is:

\[
E(2) \leq \frac{M}{60}.
\]

\subsection*{Question 2.11.1 VII}

\textbf{Answer:}\\

We will prove the given difference formulas by first stating the definitions of forward and backward differences and then applying mathematical induction.

\subsection*{1. Forward Difference Proof}

\textbf{Base Case (\( k=0 \)):}

For \( k=0 \), we have:

\[
\Delta^{0} f(x) = f(x)
\]

According to the Lagrange interpolation formula, for \( f[x_0] = f(x) \), we have:

\[
\Delta^{0} f(x) = 0! h^{0} f[x_0] \quad \text{(holds true)}
\]

\textbf{Inductive Hypothesis:}

Assume it holds for some \( k \), i.e.,

\[
\Delta^{k} f(x) = k! h^{k} f[x_{0}, x_{1}, \ldots, x_{k}]
\]

\textbf{Inductive Step:}

Now we need to prove the form of \( \Delta^{k+1} f(x) \):

\[
\Delta^{k+1} f(x) = \Delta \Delta^{k} f(x) = \Delta^{k} f(x+h) - \Delta^{k} f(x)
\]

By the inductive hypothesis, we have:

\[
\Delta^{k} f(x+h) = \Delta^{k} f(x_{1}) = k! h^{k} f[x_{1}, \ldots, x_{k}, x_{k+1}]
\]

and

\[
\Delta^{k} f(x) = k! h^{k} f[x_{0}, x_{1}, \ldots, x_{k}]
\]

Thus,

\[
\Delta^{k+1} f(x) = k! h^{k} f[x_{1}, \ldots, x_{k}, x_{k+1}] - k! h^{k} f[x_{0}, x_{1}, \ldots, x_{k}] = k! h^{k} (x_{k+1}-x_{0}) f[x_{0}, x_{1}, \ldots, x_{k+1}] = (k+1)! h^{k+1} f[x_{0}, x_{1}, \ldots, x_{k+1}]
\]

Therefore, we conclude:

\[
\Delta^{k} f(x) = k! h^{k} f[x_{0}, x_{1}, \ldots, x_{k}]
\]

holds for all non-negative integers \( k \).

\subsection*{2. Backward Difference Proof}

\textbf{Base Case (\( k=0 \)):}

For \( k=0 \), we have:

\[
\nabla^{0} f(x) = f(x)
\]

According to the Lagrange interpolation formula, for \( f[x_{0}] = f(x) \), we have:

\[
\nabla^{0} f(x) = 0! h^{0} f[x_{0}] \quad \text{(holds true)}
\]

\textbf{Inductive Hypothesis:}

Assume it holds for some \( k \), i.e.,

\[
\nabla^{k} f(x) = k! h^{k} f[x_{0}, x_{-1}, \ldots, x_{-k}]
\]

\textbf{Inductive Step:}

Now we need to prove the form of \( \nabla^{k+1} f(x) \):

\[
\nabla^{k+1} f(x) = \nabla \nabla^{k} f(x) = \nabla^{k} f(x) - \nabla^{k} f(x-h)
\]

By the inductive hypothesis, we have:

\[
\nabla^{k} f(x) = k! h^{k} f[x_{0}, x_{-1}, \ldots, x_{-k}]
\]

and

\[
\nabla^{k} f(x-h) = \Delta^{k} f(x_{-1}) = k! h^{k} f[x_{-1}, \ldots, x_{-k-1}]
\]

Thus,

\[
\nabla^{k+1} f(x) = k! h^{k} f[x_{-1}, \ldots, x_{-k}] - k! h^{k} f[x_{0}, x_{-1}, \ldots, x_{-k-1}] = k! h^{k} (x_{0}-x_{-k-1}) f[x_{0}, x_{-1}, \ldots, x_{-k-1}] = (k+1)! h^{k+1} f[x_{0}, x_{-1}, \ldots, x_{-k-1}]
\]

Therefore, we conclude:

\[
\nabla^{k} f(x) = k! h^{k} f[x_{0}, x_{-1}, \ldots, x_{-k}]
\]

holds for all non-negative integers \( k \).

\section*{Conclusion}

We have proven the following two formulas:

\[
\Delta^{k} f(x) = k! h^{k} f[x_{0}, x_{1}, \ldots, x_{k}]
\]

and

\[
\nabla^{k} f(x) = k! h^{k} f[x_{0}, x_{-1}, \ldots, x_{-k}]
\]

These two formulas are valid in the contexts of forward and backward differences.

\subsection*{Question 2.11.1 VIII}

\textbf{Answer:}\\

1. \textbf{Base Case}: When \(n=0\), the divided difference \(f[x_0, x_0]\) is the first-order divided difference, which is:

\[
f[x_0, x_0] = f'(x_0) = \frac{\partial}{\partial x_0} f[x_0]
\]

2. \textbf{Inductive Step}: Suppose the statement holds for \(n=k\), that is:

\[
\frac{\partial}{\partial x_0} f[x_0, x_1, \ldots, x_k] = f[x_0, x_0, x_1, \ldots, x_k]
\]

We need to prove that the statement also holds for \(n=k+1\). That is:

\[
\frac{\partial}{\partial x_0} f[x_0, x_1, \ldots, x_{k+1}] = f[x_0, x_0, x_1, \ldots, x_{k+1}]
\]

According to the definition of the divided difference, we can write \(f[x_0, x_1, \ldots, x_{k+1}]\) as:

\[
f[x_0, x_1, \ldots, x_{k+1}] = \frac{f[x_1, \ldots, x_{k+1}] - f[x_0, \ldots, x_k]}{x_{k+1} - x_0}
\]

Taking the partial derivative with respect to \(x_0\), we get:

\[
\frac{\partial}{\partial x_0} f[x_0, x_1, \ldots, x_{k+1}] = \frac{\frac{\partial}{\partial x_0} f[x_1, \ldots, x_{k+1}] - \frac{\partial}{\partial x_0} f[x_0, \ldots, x_k]}{x_{k+1} - x_0}
\]

By the inductive hypothesis, we can replace \(\frac{\partial}{\partial x_0} f[x_1, \ldots, x_{k+1}]\) and \(\frac{\partial}{\partial x_0} f[x_0, \ldots, x_k]\) with \(f[x_1, x_0, \ldots, x_{k+1}]\) and \(f[x_0, x_0, \ldots, x_k]\), respectively, thus obtaining:

\[
\frac{\partial}{\partial x_0} f[x_0, x_1, \ldots, x_{k+1}] = \frac{f[x_1, x_0, \ldots, x_{k+1}] - f[x_0, x_0, \ldots, x_k]}{x_{k+1} - x_0}
\]

This is precisely the definition of \(f[x_0, x_0, x_1, \ldots, x_{k+1}]\), and therefore the statement holds for \(n=k+1\).

By mathematical induction, we have proven that for all \(n\), \(\frac{\partial}{\partial x_0} f[x_0, x_1, \ldots, x_n] = f[x_0, x_0, x_1, \ldots, x_n]\).

As for the partial derivatives with respect to other variables, we can handle them similarly. For any \(x_i\), the partial derivative \(\frac{\partial}{\partial x_i} f[x_0, x_1, \ldots, x_n] = \frac{\partial}{\partial x_i} f[x_i, x_0, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n]\) can be obtained similarly to the case of $x_0$, so we get :
\[
\frac{\partial}{\partial x_i} f[x_0, x_1, \ldots, x_n] = f[x_i, x_0, x_1, \ldots, x_n]
\]

\subsection*{Question 2.11.1 IX}

\textbf{Answer:}\\

By 2.4.7, we know \( \forall p \in \tilde{P}_n \), it holds that
\[
\max_{x \in [-1,1]} \left| \frac{T_n(x)}{2^{n-1}} \right| \leq \max_{x \in [-1,1]} |p(x)|
\]
where \( T_n(x) \) denotes the \( n \)-th Chebyshev polynomial.

Chebyshev polynomials \({T_n(x)}\)'s maximum absolute value is 1. We can adapt Chebyshev polynomials to any interval \([a, b]\) by appropriate scaling and shifting.

First, we need to map the interval \([a, b]\) to \([-1, 1]\). This can be achieved through the linear transformation \(x' = \frac{a+b-2x}{a-b}\), where \(x'\) is the new variable that varies over \([-1, 1]\). Then, we use the Chebyshev polynomial \(T_n(x')\).
As \(a_0 \neq 0\), we use \(P(x) = a_0\frac{(b-a)^n}{2^n}\frac{T_n(x)}{2^{n-1}} = a_0\frac{(b-a)^nT_n(x)}{2^{2n-1}}\) as the polynomial $a_0 x^n + a_1 x^{n-1} + \cdots + a_n$ of \(\min\max_{x\in[a, b]}\left|a_0 x^n + a_1 x^{n-1} + \cdots + a_n\right|\).\\
Thus, we conclude: 
\[\min\max_{x\in[a, b]}\left|a_0 x^n + a_1 x^{n-1} + \cdots + a_n\right| = \frac{|a_0|(b-a)^n}{2^{2n-1}}\]                                                       

\subsection*{Question 2.11.1 X}

\textbf{Answer:}\\

\(T_n(x)\) assumes its extrema \(n+1\) times at the points \(x_k' = cos(\frac{k}{n}\pi)\) . Suppose the conclusion does not hold. Then it implies that
\[
\exists p \in P_n^a \quad \text{s.t.} \max_{x \in [-1,1]} |p(x)| < \max_{x \in [-1,1]} |\hat{p}_n(x)|. \quad  (*)
\]

Consider the polynomial 
\[
Q(x) = \hat{p}_n(x) - p(x).
\]
\[
Q(x_k') = \frac{(-1)^k}{T_n(a)} - p(x_k'), \quad k = 0, 1, \ldots, n.
\]

\(Q(x)\) has alternating signs at these \(n+1\) points. Hence \(Q(x)\) must have \(n\) zeros for $x \in [-1,1]$. Besides, Q(a) = 0, so \(Q(x)\) must have \(n\) zeros for $x \in R$. 
However, by the construction of \(Q(x)\), the degree of \(Q(x)\) is at most \(n\). Therefore, \(Q(x) \equiv 0\) and \(p(x) = \hat{p}_n(x)\), which implies \(\max |p(x)| = \hat{p}_n(x)\). This is a contradiction to (*). 

\subsection*{Question 2.11.1 XI}

\textbf{Answer:}\\

We start with the definition of the Bernstein basis polynomials:
\[
b_{n,k}(t) = \binom{n}{k} t^k (1-t)^{n-k}
\]
\[
b_{n,k+1}(t) = \binom{n}{k+1} t^{k+1} (1-t)^{n-k-1}
\]

We need to prove that:
\[
b_{n-1,k}(t) = \binom{n-1}{k} t^k (1-t)^{n-1-k}
\]
can be expressed as:
\[
\frac{n-k}{n} b_{n,k}(t) + \frac{k+1}{n} b_{n,k+1}(t)
\]

Substituting the expressions for \(b_{n,k}(t)\) and \(b_{n,k+1}(t)\) into the right-hand side and simplifying:
\[
\frac{n-k}{n} \binom{n}{k} t^k (1-t)^{n-k} + \frac{k+1}{n} \binom{n}{k+1} t^{k+1} (1-t)^{n-k-1}
\]

Using the properties of binomial coefficients \(\binom{n}{k} = \frac{n}{n-k} \binom{n-1}{k}\) and \(\binom{n}{k+1} = \frac{n}{k+1} \binom{n-1}{k}\), we can rewrite the above expression as:
\[
\frac{n-k}{n} \frac{n}{n-k} \binom{n-1}{k} t^k (1-t)^{n-k} + \frac{k+1}{n} \frac{n}{k+1} \binom{n-1}{k} t^{k+1} (1-t)^{n-k-1}
\]

Simplifying, we get:
\[
\binom{n-1}{k} t^k (1-t)^{n-1-k}
\]

This is exactly the definition of \(b_{n-1,k}(t)\). Therefore, we have proven that \(b_{n-1,k}(t)\) can be expressed as a linear combination of \(b_{n,k}(t)\) and \(b_{n,k+1}(t)\).

\subsection*{Question 2.11.1 XII}

\textbf{Answer:}\\

We need to compute the integral of the Bernstein basis polynomial:
\[
\int_0^1 b_{n,k}(t) \, dt = \int_0^1 \binom{n}{k} t^k (1-t)^{n-k} \, dt
\]

First, we can factor out the constant term \(\binom{n}{k}\):
\[
\binom{n}{k} \int_0^1 t^k (1-t)^{n-k} \, dt
\]

We choose \(u = t^k\) and \(dv = (1-t)^{n-k} \, dt\). Then, we have \(du = kt^{k-1} \, dt\) and \(v = -\frac{1}{n-k+1}(1-t)^{n-k+1}\).

Applying the integration by parts formula \(\int u \, dv = uv - \int v \, du\), we get:
\[
\int_0^1 t^k (1-t)^{n-k} \, dt = \left[ -\frac{t^k}{n-k+1}(1-t)^{n-k+1} \right]_0^1 + \frac{k}{n-k+1} \int_0^1 t^{k-1} (1-t)^{n-k+1} \, dt
\]

At \(t = 1\), \((1-t)^{n-k+1} = 0\), so the first term vanishes. At \(t = 0\), \(t^k = 0\), so the first term also vanishes. Therefore, we have:
\[
\int_0^1 b_{n,k}(t) \, dt = \int_0^1 \binom{n}{k}t^k (1-t)^{n-k} \, dt = \frac{k}{n-k+1} \binom{n}{k} \int_0^1 t^{k-1} (1-t)^{n-k+1} \, dt = \int_0^1 b_{n,k-1}(t) \, dt
\]

This integral is the integral of \(b_{n, k-1}(t)\). So the integral of \(b_{n, k}(t)\) is independent of 
k. In other words, for a given \(n\), the integral of \(b_{n, k}(t)\) is the same for all values of \(k\). We can conclude:
\[
\int_0^1 b_{n,k}(t) \, dt = \int_0^1 b_{n,n}(t) \, dt = \int_0^1 t^n \, dt = \left[ \frac{t^n}{n+1} \right]_0^1 = \frac{1}{n+1}
\]

% ===============================================
\section*{ \center{\normalsize {Acknowledgement}} }
None.


\printbibliography

\end{document}